Bài 6: Giải các phương...
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Bài 6: Giải các phương trình sau:
2) |
3) |
4) |
6) |
7) |
11)
12)
13)
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28 tháng 2 2021
2) Ta có: \(19-\left(x-5\right)^3=x\left(3-x^2\right)-24\left(x-6\right)\)
\(\Leftrightarrow19-\left(x^3-15x^2+75x-125\right)=3x-x^3-24x+144\)
\(\Leftrightarrow19-x^3+15x^2-75x+125=-x^3-21x+144\)
\(\Leftrightarrow-x^3+15x^2-75x+144+x^3+21x-144=0\)
\(\Leftrightarrow15x^2-54x=0\)
\(\Leftrightarrow x\left(15x-54\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\15x-54=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\15x=54\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{18}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{0;\dfrac{18}{5}\right\}\)
3) Ta có: \(x\left(5-x\right)\left(x+5\right)-4x\left(x+5\right)=2x+1-\left(2x-1\right)^2\)
\(\Leftrightarrow x\left(5-x\right)\left(5+x\right)-4x\left(x+5\right)=2x+1-\left(4x^2-4x+1\right)\)
\(\Leftrightarrow x\left(25-x^2\right)-4x^2-20x=2x+1-4x^2+4x-1\)
\(\Leftrightarrow25x-x^3-4x^2-20x-2x-1+4x^2-4x+1=0\)
\(\Leftrightarrow-x^3-x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x=0
Vậy: S={0}
1
28 tháng 2 2021
`10)x^2-11x+24=0`
`<=>x^2-3x-8x+24=0`
`<=>x(x-3)-8(x-3)=0`
`<=>(x-3)(x-8)=0`
`<=>` $\left[ \begin{array}{l}x=3\\x=8\end{array} \right.$
`8,(x+1)^3-4(x+1)=0`
`<=>(x+1)[(x+1)^2-4]=0`
`<=>(x+1)(x+1-2)(x+1+2)=0`
`<=>(x+1)(x-2)(x+3)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=-1\\x=-3\end{array} \right.$
28 tháng 2 2021
4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)
\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)
\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)
\(\Leftrightarrow-22x-60=-75x-30\)
\(\Leftrightarrow-22x+75x=-30+60\)
\(\Leftrightarrow53x=30\)
\(\Leftrightarrow x=\dfrac{30}{53}\)
Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)
5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)
\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)
\(\Leftrightarrow10x-6-21x+3=60\)
\(\Leftrightarrow-11x-3=60\)
\(\Leftrightarrow-11x=63\)
\(\Leftrightarrow x=-\dfrac{63}{11}\)
Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)
28 tháng 2 2021
`9,x^3+x^2-2=0`
`x^3-x^2+2x^2-2=0`
`<=>x^2(x-1)+2(x-1)(x+1)=0`
`<=>(x-1)(x^2+2x+2)=0`
`<=>x=1`
`14,x^2-2x+1=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1`
`15,x^3+3x^2+3x+1=0`
`<=>(x+1)^3=0`
`<=>x+1=0`
`<=>x=-1`
28 tháng 2 2021
Bài 6:
1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)
\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)
\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)
\(\Leftrightarrow-14x-9=0\)
\(\Leftrightarrow-14x=9\)
\(\Leftrightarrow x=-\dfrac{9}{14}\)
Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)
28 tháng 2 2021
`1)2x(x-5)-(x+3)^2=3x-x(5-x)`
`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`
`<=>x^2-16x-9=x^2-2x`
`<=>14x=-9`
`<=>x=-9/14`
2
28 tháng 2 2021
`x(x-1)(x+1)(x+2)=24`
`<=>[x(x+1)][(x-1)(x+2)]=24`
`<=>(x^2+x)(x^2+x-2)=24`
`<=>(x^2+x-1)^2=25`
`+)x^2+x-1=5`
`<=>x^2+x-6=0`
`<=>x^2-2x+3x-6=0`
`<=>x(x-2)+3(x-2)=0`
`<=>(x-2)(x+3)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.$
`+)x^2+x-1=-5`
`<=>x^2+x+4=0`
`<=>(x+1/2)^2+15/4=0` vô lý
Vậy `S={2,3}`
28 tháng 2 2021
`11)4x^2+4-8x=9(x-2)^2`
`<=>4(x^2-2x+1)=9(x-2)^2`
`<=>(2x-2)^2=(3x-6)^2`
`<=>(3x-6-2x+2)(3x-6+2x-2)=0`
`<=>(x-4)(5x-8)=0`
`<=>` $\left[ \begin{array}{l}x=4\\x=\dfrac{8}{5}\end{array} \right.$
Vậy `S={4,5/8}`
21 tháng 3 2021
lệnh for...to...do:
a)program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
for i:=1 to 9 do s:=s+i;
write(s);
readln;
end.
b)
program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
for i:=1 to 14 do
begin
if i mod 2=0 then
s:=s+i;
end;
write(s);
readln;
end.
c)
program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
for i:=1 to 15 do
begin
if i mod 2=1 then
s:=s+i;
end;
write(s);
readln;
end.
lệnh while...do
a)program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
i:=1;
while i<=9 do
begin
s:=s+i;
i:=i+1;
end;
write(s);
readln;
end.
b)program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
i:=1;
while i<=14 do
begin
if i mod 2=0 then
s:=s+i
else i:=i+1;
end;
write(s);
readln;
end.
c)
program tinh_tong;
uses crt;
var i,s:byte;
begin
clrscr;
s:=0;
i:=1;
while i<=15 do
begin
if i mod 2=1 then
s:=s+i
else i:=i+1;
end;
write(s);
readln;
end.
24 tháng 3 2020
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
7) Ta có: \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
mà \(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: S={-66}